On Nov 29, 6:32 am, Dan Christensen <Dan_Christen...@sympatico.ca>wrote:> On Nov 28, 2:41 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:>>>>>>>>>> > On Nov 29, 1:13 am, Dan Christensen <Dan_Christen...@sympatico.ca>> > wrote:>> > > On Nov 28, 1:35 am, Graham Cooper <grahamcoop...@gmail.com> wrote:>> > > > On Nov 28, 2:26 pm, Dan Christensen <Dan_Christen...@sympatico.ca>> > > > wrote:> > > > > On Nov 27, 8:59 pm, Graham Cooper <grahamcoop...@gmail.com> wrote:>> > > > > > There are 2 ALLs which is more complicated but you can format it as a> > > > > > SUBSET using a cartesian product of the 2 X values with a common Y.>> > > > > > isfunction(r) <- ALL(Y1) ALL(Y2) r(X,Y1)^r(X,Y2) -> Y1=Y2>> > > > > There is more to functionality than this. I may not fully understand> > > > > your unusual notation (PROLOG?), but it would seem you have left out> > > > > the requirement that FOR ALL elements of some domain set, THERE EXISTS> > > > > a unique image in a codomain set. (This is where I think quantifiers> > > > > become indispensable).>> > > > > > {(Y1,Y2)|r(X,Y1)^r(X,Y2)} C {(Y1,Y2)|r(X,Y1)^r(X,Y2)->Y1=Y2}>> > > > > [snip]>> > > > > The same comment applies... I think.>> > > > > Anyway, I am still waiting for proofs of the following:>> > > > > 1. {(x,y) | x in S, y=x} is a function mapping the set S onto itself> > > > > 2. {x | ~x in x} cannot exist> > > > > 3. {x | x=x} cannot exist>> > > > > You really need to address these fundamental results.>> > > > > For what it is worth, and from what little I know about PROLOG, it> > > > > doesn't seem to be capable of all that is required to do mathematical> > > > > proofs in general. It may be able to model some interesting and useful> > > > > aspects of predicate logic and set theory, but, for your purposes,> > > > > important pieces of the puzzle seem to be missing.>> > > > Nope, this is exactly the definition of function.>> > > > {(Y1,Y2) | r(X,Y1)^r(X,Y2)} C {(Y,Y) | r(X,Y)}>> > > > which simply guarantees only 1 Y value for any X value.>> > > This is a common mistake. According to this erroneous view, every set> > > {(x,y)} is a function for any objects x and y. The functionality of a> > > set of ordered pairs is always defined in terms of a domain and> > > codomain set. Here is a typical formal(ish) definition of a function> > > from Wiki (my comments in []'s):>> > > "A function f from X [the domain of f] to Y [the codomain of f] is a> > > subset of the Cartesian product X × Y subject to the following> > > condition: every element of X is the first component of one and only> > > one ordered pair in the subset.[3] In other words, for every x in X> > > there is exactly one element y such that the ordered pair (x, y) is> > > contained in the subset defining the function f."http://en.wikipedia.org/wiki/Function_(mathematics)#Definition>> > > Example: Let X=Y={0,1}.>> > > Then {(0,0), (1,1)} is function from X to Y, while {(0,1)} is not.>> > Semantics escapes you!>> Just how are you going to implement the above definition of a function> in your formal system without universal and existential quantifiers?> You can't.>> This looks like a major setback for your project, Herc. If you want my> opinion -- I can't imagine that your do -- but nevertheless I really> don't think PROLOG is the way to go. It is not a theorem prover. It is> just another database query language. It doesn't do generalizations.> For that, you need quantifiers.>> > {(0,1)} is a function.>> You can say {(0,1)} is a function only if you also say that the domain> of that function is {0}.>> Dan> Download my DC Proof 2.0 software athttp://www.dcproof.com>

You 2 aren't making any sense.

I converted the Quantified Defn to a SUBSET defn.

If you want a 2 parameter predicate

isFunctionOf( r , d )

instead of isFunction( r )

it's also do-able, but I have no idea what purpose isFunctionOf wouldhave.

PROLOG is just the UNIFY( f1(..) , f1(...) )

You need it to match the parameters of the LHS of the inference rule,as it happens it's also computationally complete.